A worked-example animation solving two energy efficiency problems in Arabic: an electric hoist converting 50,000 J of input into 35,000 J of useful work, and an electric crane lifting a 250 kg load 6.0 m in 8.0 s on 2.5 kW of power. It visualizes wasted energy, builds a Sankey diagram for the hoist, and calculates work, power, and efficiency percentages step by step. Useful for students practicing efficiency and power calculations in Arabic-medium physics courses.
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Scene 9: Hoist & Crane Exercises (Problems 9, 10, & 11) Problem 9 (Exercise 3 - Electric Hoist): Prompt & Motion: Electric hoist supplied with 5.0×10 4 J of electrical energy and providing 3.5×10 4 J of useful work. Voiceover & Step-by-Step Breakdown: (A) Wasted Energy: 50,000 J−35,000 J=15,000 J ( 1.5×10 4 J ). (B) Efficiency: η=35,000/50,000=0.70=70% . (C & D) Sankey Diagram & Main Wasted Energy Form: Input 50 kJ splits into 35 kJ useful and 15 kJ wasted. Main form of wasted energy is thermal energy (heat). Problems 10 & 11 (Exercise 4 - Electric Crane): Prompt & Motion: Crane lifting a 250 kg load 6.0 m vertically at constant speed in 8.0 s, supplied with electrical power at a constant rate of 2.5 kW. Voiceover & Step-by-Step Breakdown: (A) Useful Power Output: Work: W=250×9.8×6.0=14,700 J . Power: P=14,700/8.0=1,837.5 W ( 1.84 kW ). (B) Crane Efficiency: Total Input Energy (8.0 s): E in =2,500 W×8.0 s=20,000 J . Efficiency: η=14,700/20,000=0.735=73.5% . IMPORTANT LANGUAGE RULE: All on-screen text must be in Arabic EXCEPT mathematical equations, formulas, laws, symbols, variables, and numerical expressions. Keep every equation, formula, and law exactly as provided, without translating or changing it. Translate only the explanatory text, headings, labels, and descriptions into Arabic. Arabic text must use proper RTL direction and correct Arabic typography. Do not use English text anywhere else on the screen. Audience: student